A,ar,ar^2,ar^3,.....infinite number of terms

Here r < 1. So the sum converges.

Sum = a / (1 - r) = 3 => a = 3 (1 - r) ---- (1)

Series with terms being squares of the above series:

a², a² r², a²r⁴, a² r⁶, a² r⁸ ,...... infinity..

Sum of the terms = a² / ( 1 - r²) = 3

=> a² = 3 (1 - r²) --- (2)

divide (2) by (1): a = 1 + r --- (3)

(1) - (3) : 2 - 4 r = 0 => r = 1/2

hence, by (3) , a = 3/2

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If the Roots of the equation a (b-c) x² + b (c-a) x + c (a-b) = 0 are equal it means that the discriminant is 0.

Discriminant = b² (c-a)² - 4*a (b-c) * c (a-b) = 0 --(1)

If a, b, c are in AP, then: b - a = c - b and c - a = 2 (c - b) = 2 (b - a)

Substituting in (1), b² 2² (c - b)² = 4 a c (b - c)(a - b)

=> b² (c - b) = a c (a - b)

If a, b, c are in GP: b² = a c

=> (c - a)² = 4 (b - c) (a - b)

=> c² + a² + 2 ac = 4 a b - 4 b² + 4 b c

=> (c + a)² = 4 b (a + c) - 4 b²

If a,b,c are in HP then: 1/a, 1/b , 1/c are in AP.

=> 1/b - 1/a = 1/c - 1/b

=> c (a - b) = a (b - c)

Substituting in (1):

b² (c -a)² = 4 a² (b - c)²

=> b ( c - a) = + 2 a (b - c) = + 2 c (a - b) --- (2)

OR b (c - a) = - 2 a (b - c) = - 2 c (a - b) --- (3)

By (2) => (b + 2 a) c = 3 a b and 3 b c = a (b +2 c)

=> 9 a b - 6 a c = a b + 2 a c

=> 8 a b = 8 a c => b = c or a = 0

By (3): => a b + 2 a c + b c = 0 => 1/c + 2/b + 1/a = 0

and b c + a b - 2 c a = 0 => 1/a + 1/c - 2/b = 0

Hence, a,b,c are not in AP, GP or HP.